RaySix

Post

Tangent line and linear approximation

Created by @nathanedwards

at November 2nd 2023, 12:53:26 am.

AP_Calculus_AB-Questions

#derivative

#linear-approximation

#tangent-line

Question:

Let $f(x) = 4x^3 + 2x^2 - 3x + 5$ . Write the equation of the tangent line at $x=2$ using linear approximation and find the approximate value of $f(2.1)$ .

Answer:

To find the equation of the tangent line at $x=2$ , we will use linear approximation. The equation of the tangent line can be expressed as $y = mx + b$ , where $m$ is the slope of the tangent line and $b$ is the y-intercept.

The slope of the tangent line at $x=2$ is equal to the derivative of the function $f(x)$ evaluated at $x=2$ . Let's first find $f'(x)$ .

Given $f(x) = 4x^3 + 2x^2 - 3x + 5$ , we can find $f'(x)$ by differentiating each term:

f'(x) = \frac{d}{dx}(4x^3) + \frac{d}{dx}(2x^2) - \frac{d}{dx}(3x) + \frac{d}{dx}(5)

Using the power rule for differentiation, we have:

f'(x) = 12x^2 + 4x - 3

Now, let's evaluate $f'(x)$ at $x=2$ :

f'(2) = 12(2)^2 + 4(2) - 3 = 48 + 8 - 3 = 53

So, the slope of the tangent line at $x=2$ is $m = 53$ .

To find the y-intercept, we substitute $x=2$ and $m=53$ into the equation $y = mx + b$ and solve for $b$ .

y = 53x + b

f(2) = 53(2) + b

Let's find $f(2)$ by evaluating $f(x)$ at $x=2$ :

f(2) = 4(2)^3 + 2(2)^2 - 3(2) + 5 = 4(8) + 2(4) - 6 + 5 = 32 + 8 - 6 + 5 = 39

Substituting $f(2) = 39$ and $m=53$ into the equation, we have:

39 = 53(2) + b

39 = 106 + b

b = 39 - 106

b = -67

Therefore, the equation of the tangent line at $x=2$ is $y = 53x - 67$ .

Now, to find the approximate value of $f(2.1)$ , we can use this tangent line. We substitute $x=2.1$ into the equation $y = 53x - 67$ :

f(2.1) \approx 53(2.1) - 67 = 111.3 - 67 = 44.3

Hence, the approximate value of $f(2.1)$ is $44.3$ .